Rotation of points in unit circle trigonometry

A rotation of a unit-circle point follows a fixed coordinate rule.
Rotating $(x, y)$ by $180^\circ$ gives $(-x, -y)$, a reflection through the center.
Rotating $(x, y)$ by $90^\circ$ counterclockwise gives $(-y, x)$.

The figure shows a point $A$ and its $180^\circ$ rotation $B$.
The two points are diametrically opposite, connected by a line through the center.
Each coordinate of $A$ becomes its negative in $B$.

Rotating by $180^\circ$ is the same as adding $\pi$ to the angle.
So the point at angle $\frac{7\pi}{6} = \pi + \frac{\pi}{6}$ is the $180^\circ$ rotation of the point at $\frac{\pi}{6}$.
If $\frac{\pi}{6}$ gives $(x, y)$, then $\frac{7\pi}{6}$ gives $(-x, -y)$.

The new quadrant tells you the signs of the coordinates.
Angle $\frac{7\pi}{6}$ lands in Quadrant III, where both $x$ and $y$ are negative.
That matches the $(-x, -y)$ result from the $180^\circ$ rotation.

A point $(0.6, 0.8)$ lies on the unit circle. Where does it go after a $180^\circ$ rotation?
A $180^\circ$ rotation negates both coordinates: $(x, y) \to (-x, -y)$.
So it moves to $(-0.6, -0.8)$.

A point $(1, 0)$ on the unit circle is rotated $90^\circ$ counterclockwise. Where does it land?
A $90^\circ$ counterclockwise rotation sends $(x, y)$ to $(-y, x)$.
So $(1, 0)$ moves to $(0, 1)$.

The point at angle $\frac{\pi}{4}$ is $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$. What is the point at $\frac{5\pi}{4}$?
Since $\frac{5\pi}{4} = \pi + \frac{\pi}{4}$, it is the $180^\circ$ rotation, negating both coordinates.
So it is $\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$.